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Vadim26 [7]
3 years ago
10

One canned juice drink is 20​% orange​ juice; another is 55​% orange juice. How many liters of each should be mixed together in

order to get 15L that is 88​% orange​ juice?
Mathematics
1 answer:
gavmur [86]3 years ago
8 0
The answer is that it is not possible to increase the percentage beyond 55℅

even if you add 1L 22℅ to 1L 55℅ the answer would lie between the two numbers, but never exceed 55℅
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Find the distance between<br> 2-3i<br> And<br> 9+21i
Goryan [66]

2 - 3i is just another of writing (2, -3) in the cartesian plane, the 2-3i however is using the "imaginary axis" for the imaginary plane, is all however the imaginary axis is simply the equivalent of the y-axis.  Likewise 9+21i is just (9,21), so let's just use the distance formula.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{21})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[9-2]^2+[21-(-3)]^2}\implies d=\sqrt{(9-2)^2+(21+3)^2} \\\\\\ d=\sqrt{7^2+24^2}\implies d=\sqrt{49+576}\implies d=\sqrt{625}\implies d=25

3 0
3 years ago
Please answer it for me too
djyliett [7]

Answer:

C

Step-by-step explanation:

I hope i got it correct :)

6 0
3 years ago
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In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
Estimate the perimeter and the area of the shaded figure to the nearest whole number.
soldier1979 [14.2K]

Answer:

Step-by-step explanation:

The area and perimeter of the given shaded figure are respectively 33.12 unit² and 20.56 units.

Area and perimeter-based problem:

What information do we have?

Radius of semi-circle = 8 / 2 = 4 unit

Length of remain Rectangle = 8 unit

Width of remain Rectangle = 2 unit

Perimeter of shaded figure = πr + (l + 2b)

Perimeter of shaded figure = (3.14)(4) + [4 + (2)(2)]

Perimeter of shaded figure = 12.56 + 4 + 4

Perimeter of shaded figure = 20.56 units

Area of shaded figure = πr²/2 + lb/2

Area of shaded figure = (3.14)(4)²/2 + (8)(2)/2

Area of shaded figure = (3.14)(8) + 8

Area of shaded figure = 25.12 + 8

Area of shaded figure = 33.12 unit²

5 0
2 years ago
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Anwser: <br> 36÷(1-9×1-4)+1+2
kkurt [141]
So this is a BIDMAS question so first you do what’s in the brackets so you have (1-9x1-4) and you have to do the multiplication first out of that to get (1-9-4) which is -12 so then you have overall 36 / -12 + 1 + 2 so you have to do the division first so do 36/-12 to get -3 and you have -3 + 1 + 2 which is 0 so your answer is 0.
5 0
3 years ago
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